Question: Solve for $x$, ignoring any extraneous solutions: $\dfrac{x^2 + 3x}{x + 9} = \dfrac{-x + 45}{x + 9}$
Multiply both sides by $x + 9$ $ \dfrac{x^2 + 3x}{x + 9} (x + 9) = \dfrac{-x + 45}{x + 9} (x + 9)$ $ x^2 + 3x = -x + 45$ Subtract $-x + 45$ from both sides: $ x^2 + 3x - (-x + 45) = -x + 45 - (-x + 45)$ $ x^2 + 3x + x - 45 = 0$ $ x^2 + 4x - 45 = 0$ Factor the expression: $ (x - 5)(x + 9) = 0$ Therefore $x = 5$ or $x = -9$ However, the original expression is undefined when $x = -9$. Therefore, the only solution is $x = 5$.